Let $f_0(x)=x+|x-100|-|x+100|$, and for $n\geq 1$, let $f_n(x)=|f_{n-1}(x)|-1$. For how many values of $x$ is $f_{100}(x)=0$?
Explanation: For integers $n \ge 1$ and $k \ge 0,$ if $f_{n - 1}(x) = \pm k,$ then
\[f_n(x) = |f_{n - 1}(x)| - 1 = k - 1.\]This means if $f_0(x) = \pm k,$ then $f_k(x) = 0.$

Furthermore, if $f_n(x) = 0,$ then $f_{n + 1}(x) = -1,$ and $f_{n + 2}(x) = 0.$  Hence, $f_{100}(x) = 0$ if and only if $f_0(x) = 2k$ for some integer $k,$ $-50 \le k \le 50.$

We can write
\[f_0(x) = \left\{
\begin{array}{cl}
x + 200 & \text{if $x < -100$}, \\
-x & \text{if $-100 \le x < 100$}, \\
x - 200 & \text{if $x \ge 100$}.
\end{array}
\right.\][asy]
unitsize(0.01 cm);

draw((-400,-200)--(-100,100)--(100,-100)--(400,200));
draw((-400,0)--(400,0));
draw((0,-200)--(0,200));

label("$y = f_0(x)$", (400,200), E);
label("$(-100,100)$", (-100,100), N);
label("$(100,-100)$", (100,-100), S);
[/asy]

Thus, the equation $f_0(x) = \pm 100$ has two solutions, and the equation $f_0(x) = 2k$ has three solutions for $-49 \le k \le 49.$  Thus, the number of solutions to $f_{100}(x) = 0$ is $2 + 2 + 3 \cdot 99 = \boxed{301}.$